📚 IGCSE Maths Teacher Guide G-5-500: Quadratic Equations | IGCSE 数学教师用书 G-5-500：二次方程
Quadratic equations are a central topic in the IGCSE Mathematics syllabus. They appear in algebra, graphs, inequalities and real-world modelling questions. This guide covers the standard form, four solving methods, the discriminant, graphing, turning points and common exam pitfalls. By mastering these ideas, you can tackle a wide range of IGCSE questions with confidence.
二次方程是 IGCSE 数学大纲中的核心主题。它们出现在代数、图像、不等式和实际建模题中。本指南涵盖标准形式、四种求解方法、判别式、图像、顶点以及常见考试陷阱。掌握这些内容后，你就能自信地应对各类 IGCSE 题目。
1. What is a Quadratic Equation? | 什么是二次方程？
gree two. Its standard form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
二次方程是一个二次多项式方程。其标准形式为 ax² + bx + c = 0，其中 a、b、c 是常数，且 a ≠ 0。
ax² + bx + c = 0, a ≠ 0
ation becomes linear, not quadratic. Quadratic means the highest power of the variable x is 2.
表达式 ax² + bx + c 称为二次式。如果 a = 0，方程就变成线性方程，不再是二次方程。二次意味着变量 x 的最高次数为 2。
In IGCSE exams, you may be asked to write an equation in standard form, identify a, b and c, and then decide which solving method is suitable.
在 IGCSE 考试中，你可能会被要求将方程写成标准形式，指出 a、b、c，并选择适合的求解方法。
2. Solving by Factorising | 因式分解法求解
be written as a product of two linear factors. For example, x² + 5x + 6 = (x + 2)(x + 3).
当二次式可以写成两个一次因式的乘积时，因式分解法通常是最快的方法。例如，x² + 5x + 6 = (x + 2)(x + 3)。
et x = −2, and from x + 3 = 0 we get x = −3. Therefore the solutions are x = −2 or x = −3.
解 x² + 5x + 6 = 0 时，令每个因式等于零。由 x + 2 = 0 得 x = −2，由 x + 3 = 0 得 x = −3。因此解为 x = −2 或 x = −3。
Always make sure the right-hand side is 0 before factorising. If the equation is not equal to zero, rearrange it first.
在因式分解前始终确保等号右边为 0。如果方程不等于零，先移项整理。
- Check for common factors first: 2x² + 4x = 2x(x + 2).
- 先检查公因式：2x² + 4x = 2x(x + 2)。
- Do not divide both sides by x, as this may lose the solution x = 0.
- 不要两边同时除以 x，否则可能丢失 x = 0 这个解。
3. The Quadratic Formula | 求根公式
use the quadratic formula. It solves any quadratic equation of the form ax² + bx + c = 0.
当因式分解困难或无法进行时，使用求根公式。它可以求解任何形如 ax² + bx + c = 0 的二次方程。
x = (−b ± √b² − 4ac) / (2a)
two values: one using the plus sign and one using the minus sign. The expression under the square root, b² − 4ac, is called the discriminant.
符号 ± 表示你必须计算两个值：一个使用加号，一个使用减号。根号下的表达式 b² − 4ac 称为判别式。
r example, solve 2x² − 4x − 6 = 0. Here a = 2, b = −4, c = −6. Substituting gives x = (4 ± √16 + 48) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4, so x = 3 or x = −1.
例如，解 2x² − 4x − 6 = 0。这里 a = 2，b = −4，c = −6。代入得 x = (4 ± √16 + 48) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4，所以 x = 3 或 x = −1。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic as a squared binomial plus a constant. This is useful for solving, graphing and finding the vertex.
配方法将二次式改写为一个二项式的平方加上一个常数。这对求解、绘图以及求顶点都很有用。
ch is 3, and square it to get 9. Then write x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4.
以 x² + 6x + 5 为例。取 x 系数的一半，即 3，再平方得 9。然后写成 x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4。
)² − 4 = 0, so (x + 3)² = 4. Taking square roots gives x + 3 = ±2, hence x = −1 or x = −5.
解 x² + 6x + 5 = 0 时，令 (x + 3)² − 4 = 0，因此 (x + 3)² = 4。开平方得 x + 3 = ±2，所以 x = −1 或 x = −5。
When the coefficient of x² is not 1, factor it out first before completing the square.
当 x² 的系数不是 1 时，先提出该系数，再进行配方。
5. The Discriminant | 判别式
y Δ = b² − 4ac. It tells us the nature of the roots without actually solving the equation.
判别式由 Δ = b² − 4ac 给出。它无需实际求解方程即可说明根的性质。
| Discriminant | Nature of roots | 根的性质 |
| Δ > 0 | two distinct real roots | 两个不相等的实根 |
| Δ = 0 | one repeated real root | 一个重根（两个相等的实根） |
| Δ < 0 | no real roots | 没有实根 |
inant is (−4)² − 4(1)(13) = 16 − 52 = −36. Since Δ < 0, the equation has no real roots.
例如，对于 x² − 4x + 13 = 0，判别式为 (−4)² − 4(1)(13) = 16 − 52 = −36。因为 Δ < 0，该方程没有实根。
6. Graphs of Quadratic Functions | 二次函数图像
c is a parabola. If a > 0, the parabola opens upwards. If a < 0, it opens downwards.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上；如果 a < 0，开口向下。
erm c, because when x = 0, y = c. The x-intercepts are the real roots of ax² + bx + c = 0.
y 轴截距是常数项 c，因为当 x = 0 时，y = c。x 轴截距是 ax² + bx + c = 0 的实根。
symmetry is a vertical line that passes through the vertex. Its equation is x = −b / (2a).
对称轴是一条经过顶点的竖直直线。其方程为 x = −b / (2a)。
Axis of symmetry: x = −b / (2a)
Sketching a parabola in IGCSE often requires you to label the y-intercept, any x-intercepts and the vertex.
在 IGCSE 中画抛物线草图时，通常需要标出 y 轴截距、任意 x 轴截距以及顶点。
7. Turning Points and Symmetry | 顶点与对称性
The vertex of a parabola is its turning point. For a > 0, it is a minimum point. For a < 0, it is a maximum point.
抛物线的顶点是其转折点。当 a > 0 时，它是最小值点；当 a < 0 时，它是最大值点。
k, the vertex is simply (h, k). This is why completing the square is so useful for graphing.
在配方法形式 y = a(x − h)² + k 中，顶点就是 (h, k)。这就是配方法在绘图时如此有用的原因。
x = −b / (2a), then substitute this value back into the equation to find the y-coordinate.
如果给出的是标准形式，先用 x = −b / (2a) 求出顶点的 x 坐标，再将该值代回方程求出 y 坐标。
The parabola is symmetric about its axis of symmetry. This symmetry can be used to check your sketch or find missing points.
抛物线关于其对称轴对称。这种对称性可以用来检查你的草图或找出缺失的点。
8. Quadratic Inequalities | 二次不等式
To solve a quadratic inequality such as ax² + bx + c > 0 or ax² + bx + c < 0, first find the roots of the related quadratic equation.
要解形如 ax² + bx + c > 0 或 ax² + bx + c < 0 的二次不等式，首先求出相应二次方程的根。
These divide the number line into three intervals: x < 1, 1 < x < 3 and x > 3.
例如，解 x² − 4x + 3 > 0。根为 x = 1 和 x = 3。它们将数轴分成三个区间：x < 1，1 < x < 3 和 x > 3。
Test a value from each interval in the original inequality. The parabola is above the x-axis when x < 1 or x > 3, so the solution is x < 1 or x > 3.
在每个区间中取一个值代入原不等式检验。当 x < 1 或 x > 3 时，抛物线位于 x 轴上方，因此解为 x < 1 或 x > 3。
x ∈ (−∞, 1) ∪ (3, ∞)
(≥ or ≤). Strict inequalities exclude the roots; inclusive inequalities include the roots.
始终注意不等式是严格（> 或 <）还是包含（≥ 或 ≤）。严格不等式不包括根；包含不等式包括根。
9. Real-World Applications | 实际应用
e modelled by h(t) = −5t² + 20t + 1, where h is height in metres and t is time in seconds.
二次方程可以模拟许多现实情境，包括抛体运动、面积优化和利润分析。例如，向上抛出的球的高度可建模为 h(t) = −5t² + 20t + 1，其中 h 为高度（米），t 为时间（秒）。
uation. In context, negative time solutions are usually rejected because time cannot be negative.
要求球何时落地，令 h(t) = 0 并求解二次方程。在具体情境中，负时间解通常舍去，因为时间不能为负。
3 and width x, and its area is 40, then x(x + 3) = 40, which expands to x² + 3x − 40 = 0.
面积问题常常引出二次方程。例如，一个矩形场地的长为 x + 3，宽为 x，面积为 40，则 x(x + 3) = 40，展开得 x² + 3x − 40 = 0。
Always interpret your mathematical solution in the original context. Check that lengths, times and quantities are positive and realistic.
始终在原始情境中解释你的数学解。检查长度、时间和数量是否为正且符合实际。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
One of the most common IGCSE mistakes is forgetting to set the equation equal to zero before factorising. Always rearrange first.
IGCSE 中最常见的错误之一是因式分解前忘记让方程等于零。务必先移项整理。
an remove the solution x = 0. Instead, factor the x out and use the zero product property.
另一个常见错误是两边同时除以 x。这可能会去掉 x = 0 这个解。正确做法是将 x 提出来，并利用零积性质。
ic formula. Missing it gives only one root instead of two. Also, be careful with negative values of b: subtracting a negative changes the sign.
不要忘记求根公式中的 ± 号。漏掉它只会得到一个根，而不是两个。同时，注意 b 为负值的情况：减去一个负数会改变符号。
In IGCSE exams, marks are awarded for method as well as final answers. Always show your substitution, simplification and checking steps.
在 IGCSE 考试中，方法分和最终答案分都会给。始终展示你的代入、化简和检验步骤。
Finally, practise past paper questions under timed conditions. Become familiar with the syllabus formula sheet and know when to use each method: factorising, formula, completing the square or discriminant.
最后，在限时
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图